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CBSE Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 9 - MCQExams.com
CBSE
Class 11 Medical Chemistry
Some Basic Concepts Of Chemistry
Quiz 9
What mass of
N
a
C
l
would contain the same total number of ions as 245 g of
M
g
C
l
2
?
Report Question
0%
245 g
0%
225 g
0%
263 g
0%
None of these
Explanation
(
24
+
35.5
×
2
)
gm of
m
g
c
l
2
→
3
No ion
245 gm of
m
g
c
l
2
→
3
(
24
+
70
)
×
245
g
m
(
3
24
+
70
)
×
245
ion of Nacl
→
(
23
+
35.5
)
2
×
N
a
×
3
(
24
+
70
)
×
245
g
m
A
g
N
O
3
sample is 85% by mass. To prepare
12.5
m
of 0.05 molar
A
g
N
O
3
solution,
A
g
N
O
3
sample required is:
Report Question
0%
2.36 g
0%
1.25 g
0%
1.49 g
0%
none of these
Explanation
Given, percentage purity of
A
g
N
O
3
Sample
=
85
%
We are to prepare
125
m
L
solution haring molarity
=
0.05
.
Thus, moles of pure
A
g
N
O
3
needed = Volume
×
molarity
=
125
1000
×
0.05
m
o
=
6.25
×
10
−
3
m
o
l
.
weight of pure
AgN
O
3
needed
=
(
6.25
×
10
−
3
)
×
170
g
m
=
1.0625
g
m
.
Let, mass of
A
g
N
O
3
sample required =
m
gm.
Thus,
m
×
85
100
=
1.0625
∴
m
=
1.0625
×
100
85
g
m
=
1.25
g
m
So, correct option is (B)
A fresh
H
2
O
2
solution is labeled as
11.2
V
. Calculate its concentration in wt/vol percent.
Report Question
0%
3.03
0%
6.8
0%
1.7
0%
13.6
Explanation
D
e
n
s
i
t
y
=
v
11.2
∗
M
.
w
t
d
=
11.2
11.2
∗
34
=
34
w
t
v
=
34
11.2
=
3.03
According to the
C
a
C
O
3
+
H
C
l
→
C
a
C
l
2
+
C
O
2
+
H
2
O
. What mass of
C
a
C
O
3
is required to react completely with
25
ml. of
0.75
M
H
C
l
?
Report Question
0%
0.9945 g
0%
0.6645 g
0%
0.9375 g
0%
0.8555 g
Explanation
Using morality, Number of moles are calculated for
1000
m
l
or
1
L
of the solution.
1000
m
l
contains
0.75
M
H
C
l
H
C
l
=
0.75
mole.
Therefore,
25
m
l
of
0.75
M
HCl will contains
H
C
l
=
0.75
×
25
1000
=
0.01875
m
o
l
e
2
moles of
H
C
l
reacts with
=
1
m
o
l
e
of
C
a
C
O
3
Therefore,
0.01875
mole of
H
C
l
will react with
=
1
×
0.01875
2
=
0.009375
m
o
l
e
Molar mass of
C
a
C
O
3
=
100
g
Hence, the mass of
0.009375
of
C
a
C
O
3
=
n
o
.
o
f
m
o
l
e
s
×
m
o
l
a
r
m
a
s
s
=
0.9375
g
Two flasks
A
and
B
of
500
mL each are respectively filled with
O
2
and
S
O
2
at
300
K and
1
atm pressure. The flasks contain:
Report Question
0%
the same number of atoms
0%
the same number of molecules
0%
more number of moles in flask
A
as compared to flask
B
0%
the same mass of gases
Explanation
According to Avogadro's Hypothesis: Equal volumes of all gases contain equal number of atoms under similar conditions of temperature and pressure. So, the correct option will be '
A
'.
15 g of methyl alcohol is present in 100 mL of solution. If the density of solution is 0.96 g
m
L
−
1
, Calculate the mass percentage of methyl alcohol in solution.
Report Question
0%
45.23%
0%
15.66%
0%
25.36%
0%
None of these
Explanation
Mass of methyl alcohol = 15 g
Volume of the solution = 100 ml
Density =
0.96
g
m
L
−
1
So. Density =
m
a
s
s
v
o
l
u
m
e
mass = 100
×
0.90 = 96 g
Mass percentage of methyl alcohol =
m
a
s
s
o
f
m
e
t
h
y
l
a
l
c
o
h
o
l
m
a
s
s
o
f
s
o
l
u
t
i
o
n
×
100
=
1500
96
=
15.66
%
A gas cylinder was found unattended in a public place. The investigating team took the collected samples from it. The density of the gas was found to be
2.380
g
L
−
1
at
15
o
C
and pressure. Hence the molar mass of the gas is:
Report Question
0%
30
0%
71
0%
32
0%
58
Explanation
Applying ideal gas law,
P
V
=
n
R
T
P
=
w
t
M
w
t
×
V
o
l
u
m
e
×
R
T
M
w
t
=
d
P
×
R
T
=
2.380
1
×
0.0821
×
288
=
58
What is the mass of the precipitate formed when
50
mL of
16.9
% solution of
A
g
N
O
3
is mixed with
50
mL of
5.8
%
N
a
C
l
solution ?
(Ag =
107.8
, N =
14
, O =
16
, Na =
23
,
C
l
=
35.5
)
Report Question
0%
7
g
0%
14
g
0%
28
g
0%
3.5
g
Explanation
50
m
L
16.9
%
A
g
N
O
3
=
50
×
16.9
100
×
169
=
0.5
m
o
l
A
g
N
O
3
50
m
L
15.8
%
N
a
C
l
=
50
×
5.8
100
×
5.8
=
0.05
m
o
l
N
a
C
l
.
A
g
N
O
3
+
a
q
N
a
C
l
→
a
q
A
g
C
l
↓
+
N
a
N
O
3
(
a
q
)
0.05
0.05
- -
- -
0.05
0.05
molecular mass of
A
g
C
l
=143.3
mass of ppt
(
A
g
C
l
)
=
0.05
×
143.3
=
7.1
g
m
Hence, option
A
is correct.
Pehal and Ishaan were making
100
g
sugar solution with concentration by mass
20
%
and
50
%
respectively. Pehal added
20
g
of sugar in her solution whereas ishaan evaporated
20
g
of water from his solution. Now, the solutions were mixed to form a final solution, The concentration by mass of final solution is:
Report Question
0%
28
%
0%
45
%
0%
40
%
0%
60
%
Explanation
For Pehal:- a mass of sugar=
20
g
mass of water=
80
g
The new mass of sugar=
20
+
20
=
40
g
∴
Concentration of sugar=
40
120
×
100
=
33.33
%
For Ishaan:- mass of sugar= mass of water=
50
g
New mass of water=
50
−
20
=
30
g
∴
Concentration of sugar=
50
80
×
100
=
62.5
%
Concentration by mass of final solution=
N
e
w
m
a
s
s
o
f
s
u
g
a
r
i
n
P
e
h
a
l
+
M
a
s
s
o
f
s
u
g
a
r
i
n
I
s
h
a
a
n
N
e
w
m
a
s
s
o
f
s
o
l
u
t
i
o
n
i
n
P
e
h
a
l
+
N
e
w
m
a
s
s
o
f
s
o
l
u
t
i
o
n
i
n
I
s
h
a
a
n
=
40
+
50
120
+
80
=
45
%
3.49g of ammonia at STP occupies a volume of 4.48
d
m
3
calculate the molar mass of ammonia.
Report Question
0%
13.25 g/mol
0%
14.59
g
/
m
o
l
0%
17.45
g
/
m
o
l
0%
None of these
Explanation
Volume = 4.48
d
m
3
= 4.48 L
Gram Molecular Mass = 22.4 L at STP
So,
4.48 L of ammonia gas weighs = 3.49 g
22.4 L of ammonia weighs =
22.4
×
3.49
4.48
= 17. 45 g / mol
Two isotopes of an elements
Q
are
Q
97
(23.4% abundance) and
Q
94
(76.6% abundance).
Q
97
is
8.082
times heavier than
C
12
and
Q
94
is
7.833
times heavier than is
C
12
. What is the average atomic weight of the elements
Q
?
Report Question
0%
94.702
0%
78.913
0%
96.298
0%
94.695
Explanation
Average mass
=
Abundance of
Q
97
×
mass of
Q
97
×
Abundance of
Q
94
×
mass of
Q
94
100
=
23.4
×
8.082
×
12
+
76.6
×
7.833
×
12
100
=
94.702
Answer.
1
g-atom of nitrogen atom represents:
Report Question
0%
6.02
×
10
23
N
2
molecules
0%
22.4
L
of
N
2
at
S
.
T
.
P
0%
11.2
L
of
N
2
at
S
.
T
.
P
.
0%
28
g
of nitrogen
Explanation
1
g
−
a
t
o
m
of nitrogen atom = 1 mole of nitrogen atom
1
mole of nitrogen atom have
6.02
×
10
23
atoms and occupies
11.2
L
of
N
2
at STP.
A solution of ethanol in water is 10% by volume. If the solution and pure ethanol have densities of 0.9866 g/cc and 0.785 g/cc respectively. The percent by weight is nearly?
Report Question
0%
7.95%
0%
17%
0%
9.86%
0%
16.2%
Explanation
Volume of ethanol = 10 ml
Volume of solution = 100 ml
Weight of ethanol =
V
o
l
u
m
e
×
d
e
n
s
i
t
y
=
10
×
0.785
= 7.85 g
Weight of solution =
100
×
0.9866
= 98.66 g
Weight percent of ethanol =
7.85
98.66
×
100
=
7.95
%
10
g
of glucose is dissolved in
150
g
of water. The mass percantage of glucose is:
Report Question
0%
2.50
%
0%
6.67
%
0%
8.75
%
0%
10
%
Explanation
10
g
is dissolved in
150
g
of water
∴
Mass percentage of glucose=
10
150
×
100
=
6.67
%
N
2
gas is present in one litre flask at a pressure of
7.6
×
10
−
10
m
m
of
H
g
. The number of
N
2
gas molecules in the flask at
O
o
C
is?
Report Question
0%
2.68
×
10
9
0%
2.68
×
10
10
0%
1.34
×
10
28
0%
2.68
×
10
22
Explanation
To calculate the number of moles of the gas under the given condition by the relation
P
V
=
n
R
T
P
=
7.6
×
10
−
10
=
7.6
×
10
−
10
760
a
t
m
=
1
×
10
−
12
a
t
m
V
=
1
l
i
t
r
e
T
=
273
+
0
=
273
K
R
=
0.082
l
i
t
r
a
t
m
/
K
/
m
o
l
substituting the values in equation
n
=
P
V
R
T
=
1
×
10
−
12
×
1
0.082
×
273
m
o
l
e
s
Now since 1mole=
6.023
×
10
23
10
−
12
0.082
×
273
=
6.023
×
10
23
×
10
−
12
0.082
×
273
=
2.7
×
10
10
m
o
l
e
c
u
l
e
If one gram of
S
contains
x
atoms, atoms is one gram
O
will be:
Report Question
0%
x
0%
x
/
2
0%
2
3
x
0%
2
x
Explanation
1
gm of
S
=
1
32
moles
=
x
atoms
1
gm of
O
=
1
16
moles
∴
Atoms in
1
gm of
O
=
x
×
1
16
1
32
=
2
x
In a test tube, there is
18
g
of glucose
(
C
6
H
12
O
6
)
0.08
mole of glucose is taken out. Glucose left in the test tube is:
Report Question
0%
0.10
g
0%
0.02
g
0%
0.10
m
o
l
e
0%
3.60
g
Explanation
Moles of glucose=
18
180
=
0.1
moles
Moles of glucose left=
0.1
−
0.08
=
0.02
moles
∴
Mass of glucose left=
0.02
×
180
=
3.6
g
Rearrange the following in the order of increasing mass and choose the correct order (Atomic mass,
N
=
14
,
O
=
16
,
C
u
=
63
)
(I) 1 molecule of oxygen
(2) 1 atom of nitrogen
(3)
1
×
10
−
10
g
m
molecule of oxygen
(4)
1
×
10
−
10
g
m
of copper
Report Question
0%
I
I
<
I
<
I
V
<
I
I
I
0%
I
V
<
I
I
I
<
I
I
<
I
0%
I
I
<
I
I
I
<
I
<
I
V
0%
I
I
I
<
I
V
<
I
<
I
I
Explanation
(1)
1
molecule of oxygen=
O
2
∴
Mass of
O
2
=
16
×
2
N
A
=
32
g
N
A
=
32
6.022
×
10
23
=
5.3
×
10
−
23
g
(2)
1
atom of Nitrogen=
1
N
A
moles
=
1
6.022
×
10
23
moles
=
1.66
×
10
−
24
moles
∴
Mass of
1
atom of Nitrogen=
1.66
×
10
−
24
×
14
=
23.2
×
10
−
24
g
(3)
1
×
10
−
10
g
m
molecule of oxygen=
1
×
10
−
10
moles of
O
2
∴
Mass of
1
×
10
−
10
g
m
molecule of oxygen=
1
×
10
−
10
×
32
=
3.2
×
10
−
9
g
(4) Mass of copper
=
1
×
10
−
10
g
m
Comparing the masses in (1), (2), (3) and (4)
we get,
(
2
)
<
(
1
)
<
(
4
)
<
(
3
)
Therefore, answer is
(
I
I
)
<
(
I
)
<
(
I
V
)
<
(
I
I
I
)
.
The number of
g
molecule of oxygen in
6.0
×
10
24
molecules of
C
O
gas is:
Report Question
0%
1
g
molecule
0%
0.5
g
molecule
0%
5
g
molecule
0%
10
g
molecule
Explanation
Number of g molecule of
O
=
Number of mole of
O
gas
Number of moles of
C
O
=
6.022
×
10
24
6.022
×
10
23
=
10
No. of moles of oxygen atom
=
10
No. of g molecule of
O
is
10
2
=
5
g
1.5
g
m
mixture of
S
i
O
2
and
F
e
2
O
3
on very strong heating leave a residue weighting
1.46
g
m
. The reaction responsible for loss of weight is
F
e
2
O
3
(
s
)
⟶
F
e
3
O
4
(
s
)
+
O
2
(
g
)
What is the percentage by mass of
F
e
2
O
3
in original sample?
Report Question
0%
80
%
0%
20
%
0%
40
%
0%
60
%
Explanation
3
F
e
2
O
3
→
2
F
e
3
O
4
+
1
2
O
2
(
g
)
Molecular mass of
F
e
2
O
3
=
160
Molecular mass of
F
e
3
O
4
=
232
Loss in mass per
480
g
m
F
e
2
O
3
=
16
g
m
In this ques, loss of mass =
0.04
g
m
Mass of
F
e
2
O
3
=
0.04
16
×
480
=
1.20
g
m
∴
%
mass of
F
e
2
O
3
=
1.20
1.50
×
100
=
80
%
Calculate number of electrons present in
9.5
g
of
P
O
3
−
4
.
Report Question
0%
6
N
A
0%
4.7
N
A
0%
5
N
A
0%
0.1
N
A
Explanation
M
a
s
s
o
f
P
O
3
−
4
=
9.5
g
N
u
m
b
e
r
o
f
m
o
l
e
s
=
W
e
i
g
h
t
o
f
P
O
3
−
4
M
o
l
e
c
u
l
a
r
w
e
i
g
h
t
=
9.5
95
=
0.1
1
m
o
l
e
=
6.023
×
10
23
i
o
n
s
⇒
0.1
m
o
l
e
s
=
0.1
×
6.023
×
10
23
=
6.023
×
10
22
i
o
n
s
Each ion of
P
O
3
−
4
has
50
electrons
∴
Total number of electrons
=
6.023
×
10
22
×
50
=
3
×
10
24
electrons
=
6.023
×
10
22
×
100
2
=
6.023
×
10
23
×
10
2
=
(
N
A
)
5
=
5
N
A
The density of a
3
M
sodium thiosulphate
(
N
a
2
S
2
O
3
)
solution is
1.25
g
m
/
m
l
. Calculate the percentage by weight of sodium thiosulphate.
Report Question
0%
73.92
0%
65.84
0%
17.14
0%
37.92
Explanation
Percentage by weight of sodium thiosulphate solution
(
N
a
2
S
2
O
3
)
will be:
=
w
t
.
o
f
N
a
2
S
2
O
3
w
t
.
o
f
s
o
l
u
t
i
o
n
×
100
Also wt. of
4
N
a
2
S
O
3
=
Molarity
×
molecular wt.
=
3
×
150
%
by wt.
=
3
×
158
1.25
×
100
×
100
=
37.92
The mass of
50
%
(
w
/
w
)
solution of
H
C
l
required to react with
100
g
of
C
a
C
O
3
would be?
Report Question
0%
73
g
0%
100
g
0%
146
g
0%
200
g
Explanation
The reaction is:
2
H
C
l
+
C
a
C
O
3
⟶
C
a
C
l
2
+
C
O
+
3
H
2
2
molecules of
H
C
l
give
1
molecule of
C
a
C
O
3
(Molar mass)
C
a
C
O
3
=
100
g
100
g
of
C
a
C
O
3
=
1
m
o
l
e
So, Molar mass of
H
C
l
=
35.5
+
1
=
36.5
∴
1
m
o
l
e
of
H
C
l
=
36.5
g
Since, we get
2
m
o
l
e
s
of
H
C
l
from
100
g
(
1
m
o
l
e
)
of
C
a
C
O
3
Weight of
H
C
l
=
2
(
36.5
)
g
=
73
g
Equivalent wt.of a metal is
2.5
times higher than oxygen. The ratio of weight of metal to the oxide is:
Report Question
0%
0.4
0%
1.4
0%
2.4
0%
3.4
Explanation
Solution:- (A) 0.4
Given that the equivalent weight of a metal is
2.5
times higher than oxygen.
Equivalent weight of oxygen
=
8
∴
Equivalent weight of metal
=
equivalent wt. of oxygen
+
(
2.5
×
equivalent weight of oxygen)
=
8
+
2.5
×
8
=
28
As we know that
eq. wt.
=
mol. wt.
(
M
m
)
valence factor
(
n
)
∴
Molecular weight of metal
(
M
m
)
=
28
n
Formula of metal oxide will be-
M
2
O
n
Molecular weight of metal oxide
=
2
×
M
m
+
n
×
M
O
=
56
n
+
16
n
=
72
n
Ratio of weight of metal to its oxide
=
wt. of metal
wt. of metal oxide
=
28
n
72
n
=
0.389
≈
0.4
Hence the required answer is
(
A
)
0.4
.
"One gram moecule of gas at N.T.P occupies
22.4
litres." This fact was derived from?
Report Question
0%
Dalton's therory
0%
Avogadro's hypothesis
0%
Berzelius hypothesis
0%
Law of gaseous volume
Explanation
Finding relationship between gram molecule and volume of gas is an application of Avogadro's Hypothesis.
Result of this application is "One gram molecule of a gas at N.T.P. occupies 22.4 litres"
Hence, Option "B" is the correct answer.
One mole of a mixture of CO and
C
O
2
requires exactly 20 g of NaOH in solution for complete conversion of all the
C
O
2
into
N
a
2
C
O
3
. How much NaOH would it require for conversion into
N
a
2
C
O
3
, if the mixture (one mole) is completely oxidized to
C
O
2
?
Report Question
0%
60 g
0%
80 g
0%
40 g
0%
20 g
Explanation
let
x
and
y
mass of
C
O
and
C
O
2
be there
x
+
y
=
1
2
N
a
O
H
+
C
O
2
→
N
a
2
C
O
3
+
H
2
O
n
of
N
a
O
H
=
20
/
40
=
0.5
∴
0.5
/
2
mol
C
O
2
must be there
⇒
0.25
mol
C
O
2
∴
0.75
mol
C
O
If
C
O
is oxidised to
C
O
2
N
a
O
H
required
=
2
×
0.75
×
40
=
60
g
Vapour density of the equilibrium mixture of
N
O
2
and
N
2
O
4
is found to be 40 for the equilibrium:
N
2
O
4
⇌
2
N
O
2
.
Calculate the percentage of
N
O
2
in the mixture?
Report Question
0%
10%
0%
5%
0%
26.08%
0%
None of these
Explanation
N
2
O
4
⇌
2
N
O
2
t
=
0
1
t
=
t
e
q
1
−
α
2
α
Initial vapour density of N
2
O
4
=
92
2
=
46
Vapour density at equilibrium=40
46
40
=
c
+
c
α
c
46
40
=
1
+
α
46
1
+
α
=
40
⇒
46
=
40
+
40
α
⇒
α
−
6
40
=
3
20
∴
%
N
O
2
=
2
α
1
+
α
×
100
=
10
23
×
100
+
6
23
×
100
×
26.08
%
Hence, the correct option is
C
An unknown chlorohydrocarbon as
3.55
% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only, chlorine atoms present in
1
g
of chlorohydrocarbon are: (Atomic wt. of
C
l
=
35.5
u
; Avogadro constant
=
6.023
×
10
23
m
o
l
−
1
)
Report Question
0%
6.032
×
10
9
0%
6.032
×
10
23
0%
6.032
×
10
21
0%
6.032
×
10
20
Explanation
An unknown chlorohydrocarbon has
3.55
% of chlorine.
100 g of chlorohydrocarbon has
3.55
g of chlorine.
1 g of chlorohydrocarbon will have
3.55
×
1
100
=
0.0355
g
Atomic wt. of
C
l
= 35.5g/mol
Number of moles of
C
l
=
0.0355
g
3.55
g
/
m
o
l
=
0.001
m
o
l
Number of atoms of
C
l
=
0.001
m
o
l
×
6.023
×
10
23
m
o
l
−
1
=
6.023
×
10
20
400
m
g
capsule contains
100
m
g
of ferrous fumarate. The percentage of iron present in the capsule approximately is: (Mol mass of
F
e
(
C
4
H
2
O
4
)
=
170
)
Report Question
0%
8.2
%
0%
25
%
0%
16
%
0%
50
%
Explanation
Molar mass of ferrous fumarate
(
C
4
H
2
F
e
O
4
)
=
169.9
g
Atomic mass of Iron
(
F
e
)
=
55.8
∴
169.9
g
of
C
4
H
2
F
e
O
4
→
55.8
g
of
F
e
∴
100
m
g
of
C
4
H
2
F
e
O
4
contain:
=
55.8
g
169.9
g
×
100
m
g
=
32.8
m
g
Amount of
C
4
H
2
F
e
O
4
in
400
m
g
capsule
=
100
m
g
∴
Amount of iron in
400
m
g
capsule
=
32.8
m
g
Percentage of iron in
400
m
g
capusle
=
100
×
A
m
o
u
n
t
o
f
I
r
o
n
i
n
c
a
p
s
u
l
e
T
o
t
a
l
m
a
s
s
o
f
c
a
p
s
u
l
e
=
100
×
32.8
m
g
400
m
g
=
8.2
%
Phosphoric acid is widely used in carbonated beverages, detergents, toothpaste and fertilizers. The mass percentages of
H
,
P
and
O
in phosphoric acid are
H
−
3.06
P
−
31.63
%
,
O
−
65.31
%
. The atomic masses
H
=
1
,
P
=
31
and
O
=
16
.
Report Question
0%
True
0%
False
Explanation
Molecular formula of phosphoric acid is
H
3
P
O
4
.
∴
Molecular mass of
H
3
P
O
4
=
1
×
3
+
31
×
1
+
16
×
4
=
98
∴
Mass percentage of
H
=
1
×
3
98
×
100
%
=
3.06
%
∴
Mass percentage of
P
=
31
×
1
98
×
100
%
=
31.63
%
∴
Mass percentage of
O
=
16
×
4
98
×
100
%
=
65.31
%
Answer is true.
1
g
of a sample of brass om reacting with excess HCI produces
120
m
l
of
H
2
gas at STP. The percentage Zn in this sample of brass is: (At.wt.of Zn-=
65.5
g
)
Report Question
0%
32
%
0%
35
%
0%
38
%
0%
40
%
Explanation
Z
n
+
2
H
C
l
→
Z
n
C
l
2
+
H
2
Moles of
H
2
released
=
120
22400
=
0.0053
∴
Moles of
Z
n
=
0.0053
20
=
5.3
m
o
l
e
∴
%
of
Z
n
=
5.3
×
10
−
3
×
65.5
1
×
100
=
35.7
%
In
500
grams of water dissolved 2 moles of potassium sulphate the mass precentage of salt in soluton:
Report Question
0%
41
%
0%
60
%
0%
35
%
0%
48
%
Explanation
Given that ,
2
moles of potassium sulphate is dissolved in
500
gm of water
we know that ,
Molar mass of potassium sulphate
=
174
g
∴
Mass of potassium sulphate dissolved
=
2
×
174
=
348
g
Mass percentage of salt
=
Mass of solute
Mass of solution
×
100
%
=
348
500
+
348
×
100
%
=
348
848
×
100
%
=
41.03
%
=
41
%
The mass percentage composition of the elements in nitric acid are
H
=
1.59
%
,
N
=
22.22
%
,
O
=
76.19
%
respectively.
(
H
=
1
,
N
=
14
,
O
=
16
)
Report Question
0%
True
0%
False
Explanation
Molecular formula of nitric acid
=
H
N
O
3
.
∴
Molecular mass of
H
N
O
3
=
1
×
1
+
14
×
1
+
16
×
3
=
63
∴
Mass percentage of
H
=
1
×
1
63
×
100
%
=
1.59
%
Mass percentage of
N
=
14
×
1
63
×
100
%
=
1.59
%
Mass percentage of
O
=
16
×
3
63
×
100
%
=
76.19
%
Answer is true.
Calculate the mass of iron which will be converted into its oxide
(
F
e
3
O
4
)
by the action of
8
g
of steam on it. [At mass of
F
e
=
56
]
F
e
+
H
2
O
→
F
e
3
O
4
+
4
H
2
Report Question
0%
21
g
0%
12
g
0%
42
g
0%
84
g
Explanation
3
F
e
+
4
H
2
O
→
F
e
3
O
4
+
4
H
2
As per stiochiometry, Moles of
H
2
O
=Moles of fe
4
×
8
18
=
m
56
×
3
m
=
21
g
Therefore, mass of iron converted to
F
e
3
O
4
is
21
g
.
What is the mass of the precipitate formed when
50
mL of
16.9
%
(
w
/
v
)
solution of
A
g
N
O
3
is mixed with
50
mL of
5.8
%
(
w
/
v
)
N
a
C
l
solution?
(
A
g
=
107.8
,
N
=
14
,
O
=
16
,
N
a
=
23
,
C
l
=
35.5
)
Report Question
0%
7
0%
14
0%
28
0%
3.5
Explanation
A
g
N
O
3
+
N
a
C
l
→
A
g
C
l
↓
+
N
a
N
O
3
Moles of
A
g
N
O
3
=
16.9
×
50
100
170
=
0.05
moles
Moles of
N
a
C
l
=
5.8
×
50
100
58.5
=
0.05
moles
Therefore, the moles of precipitate
(
A
g
C
l
)
=
0.05
moles
Mass of precipitate formed
(
A
g
C
l
)
=
0.05
×
143.5
=
7.165
g
A solution of phosphoric acid was made by dissolving 10.0 g
H
3
P
O
4
in 100.0 mL water. The resulting volume was 104 mL. What is the , mole fraction of the solution
Report Question
0%
1
0%
2
0%
0.01
0%
3
0%
1.5
At
0
∘
C
the density of gaseous oxide at 2 bar is same as that of nitrogen at 5 bar and
0
∘
C
. The molar mass of the oxide is (assuming ideal behaviour):
Report Question
0%
70
g
/
m
o
l
0%
140
g
/
m
o
l
0%
28
g
/
m
o
l
0%
60
g
/
m
o
l
Explanation
We know Density
ρ
=
M
ρ
R
T
⟶
(
1
)
For the given data if
M
is the nolar mass of the gaseous oxide we have,
ρ
=
M
R
T
.2
or
2
=
ρ
R
T
M
⟶
(
2
)
Also for nitrogen
5
=
ρ
R
T
28
⟶
(
3
)
(
2
)
and
(
3
)
⇒
5
2
=
M
28
or
M
=
5
×
28
2
=
70
g
/
m
o
l
Calculate the mass percent of carbon in
C
2
H
5
O
H
(Molar mass of ethanol=46.068g)
Report Question
0%
13.13
%
0%
88.79
%
0%
52.14
%
0%
34.73
%
Explanation
Given : Mass of
C
2
H
5
O
H
=
46.08
g
Molecular mass
=
46.08
g
Percentage of carbon
=
Mass of carbon
Molecular mass
×
100
=
24
46.08
×
100
=
52.14
%
Four one litre flasks are separately filled with gases
O
2
,
F
2
,
C
H
4
and
C
O
2
under same conditions.
The ratio of the number of molecules in these gases are:
Report Question
0%
2
:
2
:
4
:
3
0%
1
:
1
:
1
:
1
0%
1
:
2
:
3
:
4
0%
2
:
2
:
3
:
4
Explanation
Avogadro's law: It states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.
n
1
:
n
2
:
n
3
:
n
4
=
P
V
R
T
:
P
V
R
T
:
P
V
R
T
:
P
V
R
T
=
1
:
1
:
1
:
1
So, the ratio of no. of molecules
=
1
:
1
:
1
:
1
The correct option is
B
.
According to Avogadro's law the volume of a gas will ____ as _____ if ____ are held constant.
Report Question
0%
increases, number of moles; P & T
0%
decreases, number of moles; P & T
0%
increases; T & P; number of moles
0%
decreases; P & T; number of moles
Explanation
According to Avogadro's law: Equal volume of all gases at same temperature and pressure will have same no. of molecules.
OR
For a given mass of ideal gas,
Volume
∝
Number of moles of the gas (if temperature and pressure are constant)
So, volume of the gas will increase as the number of moles if pressure (P) and temperature (T) are held constant.
A 1.50 g sample of an ore containing silver was dissolved,and all the
A
g
+
was converted to 0.125 g
A
g
2
S
. What was the percentage of silver in the ore?
Report Question
0%
14.23%
0%
8.27%
0%
10.8%
0%
7.2%
Explanation
Number of moles of
A
g
2
S
=
0.125
g
Molecular weight of
A
g
2
S
=
0.125
247.8
=
5
×
10
−
4
m
o
l
e
Mass of silver
=
5
×
10
4
m
o
l
e
×
2
×
108
=
0.108
g
∴
Percentage of silver in ore
=
0.108
1.5
×
100
=
7.2
%
To convert molality into which of the following unit of concentration require density of the solution?
Report Question
0%
Percentage weight by weight
0%
Percentage by volume
0%
Mole fraction
0%
Given all
Explanation
Percentage of volume depends on the volume of the solution which also depends on the density. So the correct option is B
The percentage by weight of:
a] C in carbon dioxide
b] Na in sodium carbonate
c] AI in aluminium nitride.
are
27
%
,
43.4
%
,
65.85
%
respectively
[C=12, O=16, H=1,Na=23, AI=27, N=14]
Report Question
0%
True
0%
False
Explanation
Solution:- (A) True
As we know that,
percentage by weight
=
weight
total weight
×
100
Therefore,
(a)percentage by weight of
C
in carbon dioxide-
Weight of carbon in
C
O
2
=
12
g
Weight of
C
O
2
=
44
g
∴
percentage by weight
=
12
44
×
100
=
27.27
%
≈
27
%
(b)percentage by weight of
N
a
in sodium carbonate-
Weight of sodium
=
23
g
Weight of sodium in
N
a
2
C
O
3
=
2
×
23
=
46
g
Weight of
N
a
2
C
O
3
=
106
g
∴
percentage by weight
=
46
106
×
100
=
43.4
%
(c)percentage by weight of
A
l
in aluminium nitride-
Weight of aluminium in
A
l
N
=
27
g
Weight of
A
l
N
=
41
g
∴
percentage by weight
=
27
41
×
100
=
65.85
%
Hence the given statement is true.
A gaseous mixture contains
C
H
4
and
C
2
H
6
in equimolecular proportion. The weight of 2.27 litres of this mixture at STP is:
Report Question
0%
3 g
0%
4.6 g
0%
1.6 g
0%
2.3 g
Explanation
Mol wt of
C
H
4
=
16
g
/
m
o
l
Mol wt of
C
2
H
6
=
30
g
/
m
o
l
Total
=
46
g
/
m
o
l
⇒
1
Mol wt of
C
H
4
=
22.4
L
1
Mol wt of
C
2
H
6
=
22.4
L
⇒
46
g
/
m
o
l
of
C
H
4
+
C
2
H
6
=
22.4
+
22.4
=
44.8
L
⇒
46
g
/
m
o
l
=
44.8
L
⇒
x
=
2.27
L
⇒
x
=
46
×
2.27
44.8
=
2.33
g
Hence, the answer is
2.33
g
.
Hence, the correct option is
D
25.5g
of
H
2
O
2
solution on decomposition gave
1.68L
of
O
2
at STP. The percentage strength by weight of the solution is:
Report Question
0%
30
0%
10
0%
20
0%
25
Explanation
Balanced equation for decomposition of
H
2
O
2
2
H
2
O
2
→
2
H
2
O
+
O
2
2mol
H
2
O
2
will produce 1 mol
O
2
At STP 1 mol
O
2
has volume = 22.4L
At STP 1.68L of
O
2
= 1.68/22.4 = 0.075 mol
O
2
Therefore the solution contained
0.075
×
2
=
0.15
mol
H
2
O
2
Molar mass
H
2
O
2
=
34
g
/
m
o
l
Mass of
H
2
O
2
decomposed
=
0.15
×
34
=
5.1
g
H
2
O
2
in 25.5g solution
100g solution contains :
100
/
25.5
×
5.1
=
20
The original solution contains
20
Hence, the correct option is
C
100
gm of an aq. solution of sugar contains
40
%
sugar by mass. How much water should be evaporated get
50
%
sugar solution by mass?
Report Question
0%
10
g
0%
20
g
0%
0.0
g
0%
40
g
Explanation
Initial condition
_
:
−
40
%
sugar (by mass),
Therefore, In
100
g
m
of an aqueous solution,
40
g
sugar and
60
g
water is present.
Final condition
_
:
−
50
%
sugar (by mass),
Therefore, In
100
g
m
of an aqueous solution,
50
g
sugar and
50
g
water is present.
Amount of water to be evaporated
=
60
−
50
=
10
g
r
a
m
s
According to percentage weight arrange the following in descending order in the earth crust ?
Report Question
0%
O
2
,
C
a
,
M
g
,
S
0%
O
2
,
S
,
M
g
,
C
a
0%
S
,
C
a
,
M
g
,
O
2
0%
C
a
,
O
2
,
M
g
,
S
Explanation
Oxygen is present in maximum concentration in earth's crust while
s
has less abundance,
So, abundance of availability of elements
O
2
>
C
a
>
M
g
>
S
.
One mole sample of
F
e
O
is heated in air until it is completely converted into
F
e
2
O
3
. The percentage increase in weight of the sample is?
[Given: Atomic mass of Fe = 56]
Report Question
0%
10.58
%
0%
20.28
%
0%
35.35
%
0%
11.11
%
Explanation
Mass of
1
mol of
F
e
O
=
(
56
+
16
)
=
72
g
F
e
O
+
3
4
O
2
⟶
1
2
F
e
2
O
3
1
mol of
F
e
O
gives
1
2
mol
F
e
2
O
3
.
∴
Mass of sample after conversion
=
1
2
×
M
F
e
2
O
3
=
1
2
×
160
=
80
g
∴
%
increase in weight
=
80
−
72
72
×
100
=
11.11
%
Hence, the correct option is
D
Aqueous urea solution is 20% by mass of solution. Calculate percentage by mass of solvent.
Report Question
0%
75%
0%
15%
0%
25%
0%
65%
Explanation
Let mass of the solution=
100
g
Given
20
% urea=
20
g
urea and
80
g
of water
∴
Mass of urea by solvent
=
20
80
×
100
=
25
%
The percentage weight by weight of
0.5
M
C
C
l
4
solution in benzene is
?
(Density of solution =
1.4
g/ml)
Report Question
0%
5.5
%
0%
8.25
%
0%
13
%
0%
20.06
%
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Practice Class 11 Medical Chemistry Quiz Questions and Answers
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